Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $n = \dfrac{q^2 + 3q + 2}{q - 9} \div \dfrac{7q + 7}{q - 9} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{q^2 + 3q + 2}{q - 9} \times \dfrac{q - 9}{7q + 7} $ First factor the quadratic. $n = \dfrac{(q + 1)(q + 2)}{q - 9} \times \dfrac{q - 9}{7q + 7} $ Then factor out any other terms. $n = \dfrac{(q + 1)(q + 2)}{q - 9} \times \dfrac{q - 9}{7(q + 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (q + 1)(q + 2) \times (q - 9) } { (q - 9) \times 7(q + 1) } $ $n = \dfrac{ (q + 1)(q + 2)(q - 9)}{ 7(q - 9)(q + 1)} $ Notice that $(q - 9)$ and $(q + 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(q + 1)}(q + 2)(q - 9)}{ 7(q - 9)\cancel{(q + 1)}} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $n = \dfrac{ \cancel{(q + 1)}(q + 2)\cancel{(q - 9)}}{ 7\cancel{(q - 9)}\cancel{(q + 1)}} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $n = \dfrac{q + 2}{7} ; \space q \neq -1 ; \space q \neq 9 $